Question: Multiply the following complex numbers: $({-3+4i}) \cdot ({3-3i})$
Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({-3+4i}) \cdot ({3-3i}) = $ $ ({-3} \cdot {3}) + ({-3} \cdot {-3}i) + ({4}i \cdot {3}) + ({4}i \cdot {-3}i) $ Then simplify the terms: $ (-9) + (9i) + (12i) + (-12 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ -9 + (9 + 12)i - 12i^2 $ After we plug in $i^2 = -1$ , the result becomes $ -9 + (9 + 12)i - (-12) $ The result is simplified: $ (-9 + 12) + (21i) = 3+21i $